
#1
Jan406, 12:44 PM

P: 224

I have a basic differential equation:
[tex]\frac{dy}{dx} = x + y, y(0) = 1[/tex] Now, when I try to solve this by making it exact [tex]\mu \frac{dy}{dx} + \mu y = \mu x[/tex] I get [tex]\mu = e^{x}[/tex] and solution [tex]x1[/tex]. This doesn't satisfy the initial condition. But when I try to solve it as a nonhomogenous equation as: [tex]\frac{dy}{dx} + y= x[/tex] I get [tex]y_p = 2e^x, y_c = x1[/tex] so [tex]y = 2e^xx1[/tex] Which seems to be a correct & full solution. What was I missing in the first try? 



#2
Jan406, 01:18 PM

P: 74

You are solving dy/dx + y = x which is wrong since the original equation is
dy/dx = x + y, or dy/dx  y = x so now u(x) = e^(x) and so on and it works out. And yes, the answer is y = 2e^x  x  1 



#3
Jan406, 01:54 PM

Mentor
P: 14,427

[tex]\mu \frac{dy}{dx}  \mu y = \mu x[/tex] [tex]y = (x+1) + c/\mu = ce^x  (1+x)[/tex] and then solved for the initial condition [itex]y(0)=1[/itex] yielding [itex]c=2[/itex] or [tex]y = 2e^x (x+1)[/tex] In this case, the solution to the homogeneous equation [itex]y^\primey=0[/itex] is [tex]y_c = ce^x[/tex] where [itex]c[/itex] is an arbitrary constant and [tex]y_p = (x+1)[/tex] is a particular solution to the inhomogeneous equation. Combining these, [tex]y = ce^x  (1+x)[/tex] which meets the initial conditions when [itex]c=2[/itex]. 



#4
Jan1906, 11:19 PM

P: 224

solution conflict!
Oh, that was just a typo. Thanks D_H, it ends up that I've forgotten the integration constant.
About naming, take it easy, I'm just a freshman! 


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